Question

我有一个日期pyspark数据框，其字符串列的格式为MM-dd-yyyy，我试图将其转换为日期列。

我试过了：

df.select(to_date(df.STRING_COLUMN).alias('new_date')).show()

我得到一串空值。有人可以帮忙吗？

Answer 1

在没有udf的情况下可以（首选？）

> from pyspark.sql.functions import unix_timestamp
> from pyspark.sql.functions import from_unixtime

> df = spark.createDataFrame([("11/25/1991",), ("11/24/1991",), ("11/30/1991",)], ['date_str'])

> df2 = df.select('date_str', from_unixtime(unix_timestamp('date_str', 'MM/dd/yyy')).alias('date'))

> df2

DataFrame[date_str: string, date: timestamp]

> df2.show()

+----------+--------------------+
|  date_str|                date|
+----------+--------------------+
|11/25/1991|1991-11-25 00:00:...|
|11/24/1991|1991-11-24 00:00:...|
|11/30/1991|1991-11-30 00:00:...|
+----------+--------------------+

更新（1/10/2018）：

对于Spark 2.2+，最好的方法是使用to_date或to_timestamp函数，它们都支持format参数。来自文档：

>>> df = spark.createDataFrame([('1997-02-28 10:30:00',)], ['t'])
>>> df.select(to_timestamp(df.t, 'yyyy-MM-dd HH:mm:ss').alias('dt')).collect()
[Row(dt=datetime.datetime(1997, 2, 28, 10, 30))]

Answer 2

from datetime import datetime
from pyspark.sql.functions import col, udf
from pyspark.sql.types import DateType



# Creation of a dummy dataframe:
df1 = sqlContext.createDataFrame([("11/25/1991","11/24/1991","11/30/1991"), 
                            ("11/25/1391","11/24/1992","11/30/1992")], schema=['first', 'second', 'third'])

# Setting an user define function:
# This function converts the string cell into a date:
func =  udf (lambda x: datetime.strptime(x, '%m/%d/%Y'), DateType())

df = df1.withColumn('test', func(col('first')))

df.show()

df.printSchema()

这是输出：

+----------+----------+----------+----------+
|     first|    second|     third|      test|
+----------+----------+----------+----------+
|11/25/1991|11/24/1991|11/30/1991|1991-01-25|
|11/25/1391|11/24/1992|11/30/1992|1391-01-17|
+----------+----------+----------+----------+

root
 |-- first: string (nullable = true)
 |-- second: string (nullable = true)
 |-- third: string (nullable = true)
 |-- test: date (nullable = true)

Answer 3

strptime（）方法对我不起作用。我使用cast：

获得另一个更清洁的解决方案

from pyspark.sql.types import DateType
spark_df1 = spark_df.withColumn("record_date",spark_df['order_submitted_date'].cast(DateType()))
#below is the result
spark_df1.select('order_submitted_date','record_date').show(10,False)

+---------------------+-----------+
|order_submitted_date |record_date|
+---------------------+-----------+
|2015-08-19 12:54:16.0|2015-08-19 |
|2016-04-14 13:55:50.0|2016-04-14 |
|2013-10-11 18:23:36.0|2013-10-11 |
|2015-08-19 20:18:55.0|2015-08-19 |
|2015-08-20 12:07:40.0|2015-08-20 |
|2013-10-11 21:24:12.0|2013-10-11 |
|2013-10-11 23:29:28.0|2013-10-11 |
|2015-08-20 16:59:35.0|2015-08-20 |
|2015-08-20 17:32:03.0|2015-08-20 |
|2016-04-13 16:56:21.0|2016-04-13 |

Answer 4

在接受的答案的更新中，您没有看到to_date函数的示例，因此使用该函数的另一种解决方案是：

from pyspark.sql import functions as F

df = df.withColumn(
            'new_date',
                F.to_date(
                    F.unix_timestamp('STRINGCOLUMN', 'MM-dd-yyyy').cast('timestamp'))

Answer 5

答案可能不多，所以想分享我的代码可以帮助某人

from pyspark.sql import SparkSession
from pyspark.sql.functions import to_date

spark = SparkSession.builder.appName("Python Spark SQL basic example")\
    .config("spark.some.config.option", "some-value").getOrCreate()


df = spark.createDataFrame([('2019-06-22',)], ['t'])
df1 = df.select(to_date(df.t, 'yyyy-MM-dd').alias('dt'))
print df1
print df1.show()

输出

DataFrame[dt: date]
+----------+
|        dt|
+----------+
|2019-06-22|
+----------+

上面的代码要转换为日期，如果要转换日期时间，请使用to_timestamp。让我知道是否有任何疑问。

Answer 6

尝试一下：

df = spark.createDataFrame([('2018-07-27 10:30:00',)], ['Date_col'])
df.select(from_unixtime(unix_timestamp(df.Date_col, 'yyyy-MM-dd HH:mm:ss')).alias('dt_col'))
df.show()
+-------------------+  
|           Date_col|  
+-------------------+  
|2018-07-27 10:30:00|  
+-------------------+

将pyspark字符串转换为日期格式

6 个答案: